Let $\Omega\subset\mathbb{R}^d$ be open, bounded and simply connected with smooth boundary $\partial\Omega$. Define $\mathcal{H}:=\{u\in H^1(\Omega)~|~\Delta u\in L^2(\Omega)\}$ with norm $$\|u\|_\mathcal{H}^2:=\|u\|_{H^1(\Omega)}^2+\|\Delta u\|_{L^2(\Omega)}^2.$$ I want to show that $H^2(\Omega)$ is dense in $\mathcal{H}$ under $\|\cdot\|_\mathcal{H}$.
This means, given $u\in\mathcal{H}$ and $\varepsilon>0$ I need to find a $u_0\in H^2(\Omega)$ such that $$\|u-u_0\|_{H^1(\Omega)}^2+\|\Delta u-\Delta u_0\|_{L^2(\Omega)}^2<\varepsilon.$$
Alternatively it means that given a sequence $u_n\in{H}^2(\Omega)$ that is Cauchy in $\|\cdot\|_\mathcal{H}$ I need to show that its limit $u$ is in $\mathcal{H}$.
So far I've attempted reasoning as follows:
If $u_n$ is $\mathcal{H}$-Cauchy then also it is $H^1(\Omega)$-Cauchy, i.e., $$\|u_n-u_m\|_{H^1(\Omega)}\rightarrow0\qquad(n,m\rightarrow\infty).$$ Hence by completeness there is an $H^1(\Omega)$-limit $u$: $$\|u_n-u\|_{H^1(\Omega)}\rightarrow0\qquad(n\rightarrow\infty).$$ The issue here is that $\Delta u$ is not necessarily defined. One may of course remark that by the same reasoning in the complete space $L^2(\Omega)$ the sequence $\Delta u_n$ has a limit, say, $y$, in $L^2(\Omega)$. But there is no immediate link between $y$ and $u$.
Could one attempt to solve the Dirichlet problem $$\left\{\begin{array}{rclcc} \Delta f &=&y&\text{in}&\Omega\\ f&=&u|_{\partial\Omega}&\text{on}&\partial\Omega \end{array}\right.\quad?$$ This seems like an incredibly convoluted way to do it, when I feel it should be extremely simple---$H^2(\Omega)$ is trivially $H^1$-dense in $H^1(\Omega)$.